Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07
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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Splitting a stationary set concentrated on regular cardinals

Statement

In ZFC, if κ is regular uncountable and S is a stationary subset of the regular uncountable cardinals below κ, then S has a partition into κ stationary sets.

Facts & Assumptions

[F1]

Removing the trace preserves a stationary remainder: Removing its trace from a stationary set preserves stationarity.

[F2]

Clubs are ranges of normal enumerations: Clubs on regular uncountable cardinals have normal increasing enumerations of full cardinal length.

[F3]

The diagonal intersection of clubs is club: The diagonal intersection of kappa clubs is club.

[F4]

Closure points form a club: Every self-map of kappa has a club of closure points.

[F5]

Unboundedly many stationary fibres yield a partition: Stationary tails of a regressive map yield kappa stationary pieces.

Proof

Given: The objects and hypotheses in the statement.

1.1

Let T=STr(S), stationary. For each αT use AC to choose a club Dαα disjoint from Sα, and let cα:αDα be its normal enumeration. Such clubs exist since alpha is regular uncountable and is not in the trace.

F1F2
2.1

For each coordinate ξ<κ consider the domain Tξ={αT:ξ<α}. Suppose no coordinate has stationary sets {αTξ:cα(ξ)b} for every b<κ. Choose a failing threshold bξ and avoiding club Cξ, so cα(ξ)<bξ whenever αTξCξ.

step 1.1
3.1

Let D=ξ<κCξ and let E be the club of closure points of ξbξ. Choose αTE and then γTD above alpha. For every ξ<α, diagonal membership of gamma and closure at alpha give cγ(ξ)<bξ<α. Since alpha is a nonzero limit, continuity gives cγ(α)α. Strict increase implies cγ(ξ)ξ by ordinal induction, so in fact cγ(α)=α. This contradicts DγS=, since αTS.

F2F3F4step 2.1
4.1

Consequently some ξ has all stationary tail domains. Its coordinate map on Tξ is regressive and the fibre lemma partitions Tξ into kappa stationary sets. Add STξ to one piece; this preserves stationarity and disjointness and gives the desired partition of S.

F5step 3.1

Depends on

Used by

Dependency tree · two levels

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Sources